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### Gaussian Elimination

Chemical Engineering Majors

Author(s): Autar Kaw

http://numericalmethods.eng.usf.edu

Transforming Numerical Methods Education for STEM Undergraduates

Naïve Gauss Eliminationhttp://numericalmethods.eng.usf.edu

Naïve Gaussian Elimination

A method to solve simultaneous linear equations of the form [A][X]=[C]

Two steps

1. Forward Elimination

2. Back Substitution

Forward Elimination

The goal of forward elimination is to transform the coefficient matrix into an upper triangular matrix

Forward Elimination

A set of n equations and n unknowns

. .

. .

. .

(n-1) steps of forward elimination

Forward Elimination

Subtract the result from Equation 2.

−

_________________________________________________

or

Forward Elimination

Repeat this procedure for the remaining equations to reduce the set of equations as

. . .

. . .

. . .

End of Step 1

Forward Elimination

Step 2

Repeat the same procedure for the 3rd term of Equation 3.

. .

. .

. .

End of Step 2

Forward Elimination

At the end of (n-1) Forward Elimination steps, the system of equations will look like

. .

. .

. .

End of Step (n-1)

Back Substitution

Solve each equation starting from the last equation

Example of a system of 3 equations

Back Substitution

Start with the last equation because it has only one unknown

http://numericalmethods.eng.usf.edu

Naïve Gauss EliminationExamplehttp://numericalmethods.eng.usf.edu

Example: Liquid-Liquid Extraction

A liquid-liquid extraction process conducted in the Electrochemical Materials Laboratory involved the extraction of nickel from the aqueous phase into an organic phase. A typical set of experimental data from the laboratory is given below:

Assuming g is the amount of Ni in organic phase and a is the amount of Ni in the aqueous phase, the quadratic interpolant that estimates g is given by

Example: Liquid-Liquid Extraction

The solution for the unknowns x1, x2, and x3 is given by

Find the values of x1, x2, and x3 using Naïve Gauss Elimination. Estimate the amount of nickel in organic phase when 2.3 g/l is in the aqueous phase using quadratic interpolation.

Example: Liquid-Liquid Extraction

Forward Elimination: Step 2

Yields

This is now ready for Back Substitution

Example: Liquid-Liquid Extraction

Back Substitution: Solve for x3 using the third equation

Example: Liquid-Liquid Extraction

Back Substitution: Solve for x2 using the second equation

Example: Liquid-Liquid Extraction

Back Substitution:Solve for x1 using the first equation

Example: Liquid-Liquid Extraction

The solution vector is

The polynomial that passes through the three data points is then

Where g is the amount of nickel in the organic phase and a is the amount of in the aqueous phase.

Example: Liquid-Liquid Extraction

When 2.3 g/l is in the aqueous phase, using quadratic interpolation, the estimated amount of nickel in the organic phase is

http://numericalmethods.eng.usf.edu

Naïve Gauss EliminationPitfallshttp://numericalmethods.eng.usf.edu

Is division by zero an issue here? YES

Division by zero is a possibility at any step of forward elimination

Pitfall#2. Large Round-off Errors

Exact Solution

Pitfall#2. Large Round-off Errors

Solve it on a computer using 6 significant digits with chopping

Pitfall#2. Large Round-off Errors

Solve it on a computer using 5 significant digits with chopping

Is there a way to reduce the round off error?

Avoiding Pitfalls

- Increase the number of significant digits
- Decreases round-off error
- Does not avoid division by zero

Avoiding Pitfalls

- Gaussian Elimination with Partial Pivoting
- Avoids division by zero
- Reduces round off error

http://numericalmethods.eng.usf.edu

Gauss Elimination with Partial Pivotinghttp://numericalmethods.eng.usf.edu

Pitfalls of Naïve Gauss Elimination

- Possible division by zero
- Large round-off errors

Avoiding Pitfalls

- Increase the number of significant digits
- Decreases round-off error
- Does not avoid division by zero

Avoiding Pitfalls

- Gaussian Elimination with Partial Pivoting
- Avoids division by zero
- Reduces round off error

What is Different About Partial Pivoting?

At the beginning of the kth step of forward elimination, find the maximum of

If the maximum of the values is

in the p th row,

then switch rows p and k.

Matrix Form at Beginning of 2nd Step of Forward Elimination

Example (2nd step of FE)

Which two rows would you switch?

Example (2nd step of FE)

Switched Rows

Gaussian Elimination with Partial Pivoting

A method to solve simultaneous linear equations of the form [A][X]=[C]

Two steps

1. Forward Elimination

2. Back Substitution

Forward Elimination

Same as naïve Gauss elimination method except that we switch rows before each of the (n-1) steps of forward elimination.

Example: Matrix Form at Beginning of 2nd Step of Forward Elimination

http://numericalmethods.eng.usf.edu

Gauss Elimination with Partial PivotingExamplehttp://numericalmethods.eng.usf.edu

Example 2

Solve the following set of equations by Gaussian elimination with partial pivoting

Example 2 Cont.

- Forward Elimination
- Back Substitution

Number of Steps of Forward Elimination

Number of steps of forward elimination is (n-1)=(3-1)=2

Forward Elimination: Step 1

Examine absolute values of first column, first row

and below.

- Largest absolute value is 144 and exists in row 3.
- Switch row 1 and row 3.

Forward Elimination: Step 1 (cont.)

Divide Equation 1 by 144 and

multiply it by 64, .

.

Subtract the result from Equation 2

Substitute new equation for Equation 2

Forward Elimination: Step 1 (cont.)

Divide Equation 1 by 144 and

multiply it by 25, .

.

Subtract the result from Equation 3

Substitute new equation for Equation 3

Forward Elimination: Step 2

Examine absolute values of second column, second row

and below.

- Largest absolute value is 2.917 and exists in row 3.
- Switch row 2 and row 3.

Forward Elimination: Step 2 (cont.)

Divide Equation 2 by 2.917 and

multiply it by 2.667,

.

Subtract the result from Equation 3

Substitute new equation for Equation 3

Back Substitution

Solving for a3

Back Substitution (cont.)

Solving for a2

Back Substitution (cont.)

Solving for a1

Gauss Elimination with Partial PivotingAnother Examplehttp://numericalmethods.eng.usf.edu

Partial Pivoting: Example

Consider the system of equations

In matrix form

=

Solve using Gaussian Elimination with Partial Pivoting using five significant digits with chopping

Partial Pivoting: Example

Forward Elimination: Step 1

Examining the values of the first column

|10|, |-3|, and |5| or 10, 3, and 5

The largest absolute value is 10, which means, to follow the rules of Partial Pivoting, we switch row1 with row1.

Performing Forward Elimination

Partial Pivoting: Example

Forward Elimination: Step 2

Examining the values of the first column

|-0.001| and |2.5| or 0.0001 and 2.5

The largest absolute value is 2.5, so row 2 is switched with row 3

Performing the row swap

Partial Pivoting: Example

Forward Elimination: Step 2

Performing the Forward Elimination results in:

Partial Pivoting: Example

Compare the calculated and exact solution

The fact that they are equal is coincidence, but it does illustrate the advantage of Partial Pivoting

http://numericalmethods.eng.usf.edu

Determinant of a Square MatrixUsing Naïve Gauss EliminationExamplehttp://numericalmethods.eng.usf.edu

Theorem of Determinants

If a multiple of one row of [A]nxn is added or subtracted to another row of [A]nxn to result in [B]nxn then det(A)=det(B)

Theorem of Determinants

The determinant of an upper triangular matrix [A]nxn is given by

Forward Elimination of a Square Matrix

Using forward elimination to transform [A]nxn to an upper triangular matrix, [U]nxn.

Example

Using naïve Gaussian elimination find the determinant of the following square matrix.

Forward Elimination: Step 1

Divide Equation 1 by 25 and

multiply it by 64, .

.

Subtract the result from Equation 2

Substitute new equation for Equation 2

Forward Elimination: Step 1 (cont.)

Divide Equation 1 by 25 and

multiply it by 144, .

.

Subtract the result from Equation 3

Substitute new equation for Equation 3

Forward Elimination: Step 2

Divide Equation 2 by −4.8

and multiply it by −16.8,

.

.

Subtract the result from Equation 3

Substitute new equation for Equation 3

Summary

- Forward Elimination
- Back Substitution
- Pitfalls
- Improvements
- Partial Pivoting
- Determinant of a Matrix

Additional Resources

For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit

http://numericalmethods.eng.usf.edu/topics/gaussian_elimination.html

http://numericalmethods.eng.usf.edu

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